Step 1: Frame the problem as excess kinetic energy over the bare minimum needed to escape.
By definition, the escape speed $V$ carries exactly enough kinetic energy per unit mass to cancel the gravitational binding energy at the surface, $\frac{1}{2}V^2 = \frac{GM}{R}$. Anything launched faster than $V$ still has leftover kinetic energy once it reaches infinity, since the binding energy has already been fully paid off.
Step 2: Compute the kinetic energy (per unit mass) supplied at launch.
\[
KE_{\text{launch}} = \frac{1}{2}(4V)^2 = 8V^2
\]
Step 3: Subtract exactly the binding energy, which costs $\frac{1}{2}V^2$ per unit mass, to get what survives at infinity.
\[
\frac{1}{2}v_\infty^2 = 8V^2 - \frac{1}{2}V^2 = \frac{15}{2}V^2
\]
Step 4: Solve for $v_\infty$.
\[
v_\infty^2 = 15V^2 \implies v_\infty = \sqrt{15}\,V
\]
Step 5: Conclusion.
\[
\boxed{\sqrt{15}\,V}
\]